Question

The Primary and secondary windings of a 500 KVA transformer have resistance of 0.42 Ω and 0.0019Ω

respectively. The Primary and secondary Voltage are 11000V/400V Respectively and the core loss is 2.9KW,

Assuming the Power factor of the load to be 0.8. Calculate the efficiency on (1) Full Load (2) Half Load​

Answer 1

Answer:

Answer:

0.32 and 0.006 ampere.

Explanation:

If we take the voltage of the primary coil and the secondary coil to be Vs and Vp respectively and the current and the number of turns in the secondary and primary coil be Is and Ip with Ns and Np respectively. So, Vs = (Ns/Np)Vp when Vs = Is*R = 15*Is. So, 15*Is = 10/500 * Vp. Taking the Vp as 240V we will get that Is=0.32A.

While we also know that the Is = (Np/Ns)Ip or 0.32 = 500/10*Ip which on solving we will get the value of Ip as 0.006A.